152.--ANOTHER JOINER'S PROBLEM.

[Illustration]

THE point was to find a general rule for forming a perfect square out of

another square combined with a "right-angled isosceles triangle." The

triangle to which geometricians give this high-sounding name is, of

course, nothing more or less than half a square that has been divided

from corner to corner.

The precise relative proportions of the square and triangle are of no

consequence whatever. It is only necessary to cut the wood or material

into five pieces.

Suppose our original square to be ACLF in the above diagram and our

triangle to be the shaded portion CED. Now, we first find half the

length of the long side of the triangle (CD) and measure off this length

at AB. Then we place the triangle in its present position against the

square and make two cuts--one from B to F, and the other from B to E.

Strange as it may seem, that is all that is necessary! If we now remove

the pieces G, H, and M to their new places, as shown in the diagram, we

get the perfect square BEKF.

Take any two square pieces of paper, of different sizes but perfect

squares, and cut the smaller one in half from corner to corner. Now

proceed in the manner shown, and you will find that the two pieces may

be combined to form a larger square by making these two simple cuts, and

that no piece will be required to be turned over.

The remark that the triangle might be "a little larger or a good deal

smaller in proportion" was intended to bar cases where area of triangle

is greater than area of square. In such cases six pieces are necessary,

and if triangle and square are of equal area there is an obvious

solution in three pieces, by simply cutting the square in half

diagonally.

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